Alright, this one involves a fair bit of complicated calculus, so I'm assuming you have the basics down:

First, some definitions. A normal line is a line that is perpendicular to another function at a given point. That means that the slope of the normal line is the negative reciprocal of the derivate at a given point.

We can find the slope of the given normal line (2x+9y+k=0): slope of the normal is -2/9

This means that the derivative of 3x^(2)-y^(2)=23 at some point is equal to the negative reciprocal of (-2/9), or (9/2). Let us first find the general derivative of 3x^(2)-y^(2)=23. By doing implicit differentiation, we get dy/dx=3x/y .

So we know that 3x/y=9/2. If we do some cross multiplication, it follows that y=(2/3)x. Now we can actually substitute for why into our polynomial and solve for x. By substitute I mean do this: 3x^(2)-((2/3)x)^(2)=23 [substituting (2/3)x in for y]. When we solve this for x, we get: x=+3, -3. Because y=(2/3)x, y=+2,-2.

So now we have the two points where the derivative of our polynomial has a slope of (9/2): points (3, 2) and (-3,-2)

Now that we have these points, plug the values back into 2x+9y+k=0:
2(3)+9(2)+k=0, k=-24
2(-3)+9(-2)+k=0, k=24